Quadratic Equations and Expressions
5.0 Graph of quadratic expression
5.0 Graph of quadratic expression
The general quadratic expression is represented as $$\begin{equation} \begin{aligned} f(x) = a{x^2} + bx + c \\ \Rightarrow y = a{x^2} + bx + c\quad (a \ne 0) \\ \Rightarrow y = a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{D}{{4{a^2}}}} \right] \\ \Rightarrow \left( {y + \frac{D}{{4a}}} \right) = a{\left( {x + \frac{b}{{2a}}} \right)^2} \\\end{aligned} \end{equation} $$
Now, let $\left( {y + \frac{D}{{4a}}} \right) = Y\quad {\text{and }}\left( {x + \frac{b}{{2a}}} \right) = X$ $$\begin{equation} \begin{aligned} \therefore Y = a{X^2} \\ \Rightarrow {X^2} = \frac{Y}{a} \\\end{aligned} \end{equation} $$
which is the equation of parabola.
- Axis of parabola is $X=0$ or $x + \frac{b}{{2a}} = 0$ i.e., parallel to $y$-axis.
- The co-ordinate of vertex are $\left( { - \frac{b}{{2a}}, - \frac{D}{{4a}}} \right)$
Depending on the sign of $a$ and ${b^2} - 4ac$, $f(x)$ may be positive, negative or zero. This gives rise to the following cases:
(i) If $a>0$ and ${b^2} - 4ac<0$
$ \Rightarrow f(x) > 0{\text{ }}\forall {\text{ }}x \in R$
In this case the parabola always remains above the $x$-axis and
remains concave upwards.
(ii) If $a>0$ and ${b^2} - 4ac=0$
$ \Rightarrow f(x) \geqslant 0{\text{ }}\forall {\text{ }}x \in R$
In this case the parabola touches the $x$-axis at one point and remains concave upwards.
(iii) If $a>0$ and ${b^2} - 4ac>0$
Let $f(x)=0$ have two real roots $\alpha $ and $\beta $ $(\alpha < \beta )$. Then
$\begin{equation} \begin{aligned} \Rightarrow f(x) > 0{\text{ }}\forall {\text{ }}x \in ( - \infty ,\alpha ) \cup (\beta ,\infty ) \\ \Rightarrow f(x) < 0{\text{ }}\forall {\text{ }}x \in (\alpha ,\beta ) \\\end{aligned} \end{equation} $
(iv) If $a<0$ and ${b^2} - 4ac<0$
$ \Rightarrow f(x) < 0{\text{ }}\forall {\text{ }}x \in R$
In this case the parabola always remains below the $x$-axis and remains concave downwards.
(v) If $a<0$ and ${b^2} - 4ac=0$
$ \Rightarrow f(x) \leqslant 0{\text{ }}\forall {\text{ }}x \in R$
In this case the parabola touches the $x$-axis, lies below the $x$-axis and remains concave downwards.
(vi) If $a<0$ and ${b^2} - 4ac>0$
Let $f(x)=0$ have two real roots $\alpha $ and $\beta $ $(\alpha < \beta )$. Then
$\begin{equation} \begin{aligned} \Rightarrow f(x) < 0{\text{ }}\forall {\text{ }}x \in ( - \infty ,\alpha ) \cup (\beta ,\infty ) \\ \Rightarrow f(x) > 0{\text{ }}\forall {\text{ }}x \in (\alpha ,\beta ) \\\end{aligned} \end{equation} $
Question 5. If $a{x^2} - bx + 5 = 0$ does not have two distinct real roots, then find the minimum value of $5a+b$.
Solution: Let $f(x)=a{x^2} - bx + 5 = 0$. Since $f(x)=0$ does not have two distinct real roots, we have either $f(x) \geqslant 0{\text{ }}\forall {\text{ }}x \in R$ or $f(x) \leqslant 0{\text{ }}\forall {\text{ }}x \in R$.
But $f(0)=5>0$, so $f(x)0{\text{ }}\forall {\text{ }}x \in R$
In particular, $f( - 5) \geqslant 0$ i.e., $$\begin{equation} \begin{aligned} 25a + 5b + 5 \geqslant 0 \\ 5a + b \geqslant - 1 \\\end{aligned} \end{equation} $$
Hence, the least value of $5a+b$ is $-1$.
